Question

find the remainder when p(x)=9x^3-3x^2+14x-3 is divided by g(x)=(3x-1) using remainder theorem
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Answer 1

Solution :-

Remainder Theorem :- The remainder theorem states that If you divide a polynomial f(x) by (x - h), then the remainder is f(h).

→ g(x) = 3x - 1 = 0

→ 3x = 1

→ x = (1/3)

Now,

→ p(x)=9x^3-3x^2+14x-3

→ p(1/3) = 9(1/3)³ - 3(1/3)² + 14(1/3) - 3

→ p(1/3) = 9(1/27) - 3(1/9) + (14/3) - 3

→ p(1/3) = (1/3) - (1/3) + (14 - 9)/3

→ p(1/3) = (5/3) (Ans.)

Hence, the remainder will be (5/3) .

Answer 2

$\bf \underline \pink{Answer : - }$

Reminder Theorem:-

• If you divide a polynomial f(x) by (x - h), then the remainder is f(h). The theorem states that our remainder equals f(h).

• Therefore, we do not need to use long division, but just need to evaluate the polynomial when x = h to find the remainder.

Solution:-

$\bf \: p(x) = {9x}^{3} - {3x}^{2} + 14x - 3 \\ \\ \bf \: g(x) = 3x - 1 \\ \\ \bf \: g(x) = 0 \\ \\ \bf \: 3x - 1 = 0 \\ \\ \bf \: x = \frac{1}{3} \: \: \: \\ \\ \\ now$

$\bf \implies \: p \left( \dfrac{1}{3} \right) = {9 \: ( \dfrac{1}{3} } )^{3} - 3( { \frac{1}{3} })^{2} + 14( \frac{1}{3} ) - 3 \\ \\ \bf \implies \: p \left( \dfrac{1}{3} \right) = \frac{1}{3} - \frac{1}{3} + \frac{14}{3} - 3 \\ \\ \bf \implies \: p \left( \dfrac{1}{3} \right) = \frac{14}{3} - 3 \\ \\ \bf \implies \: p \left( \dfrac{1}{3} \right) = \frac{14 - 9}{3} \\ \\ \bf \implies \: p \left( \dfrac{1}{3} \right) = \frac{5}{3}$